//给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。 
//
// 进阶：你能尝试使用一趟扫描实现吗？ 
//
// 
//
// 示例 1： 
//
// 
//输入：head = [1,2,3,4,5], n = 2
//输出：[1,2,3,5]
// 
//
// 示例 2： 
//
// 
//输入：head = [1], n = 1
//输出：[]
// 
//
// 示例 3： 
//
// 
//输入：head = [1,2], n = 1
//输出：[1]
// 
//
// 
//
// 提示： 
//
// 
// 链表中结点的数目为 sz 
// 1 <= sz <= 30 
// 0 <= Node.val <= 100 
// 1 <= n <= sz 
// 
// Related Topics 链表 双指针 
// 👍 1626 👎 0

package com.cute.leetcode.editor.cn;

import java.util.List;

public class RemoveNthNodeFromEndOfList {
    public static void main(String[] args) {
        Solution solution = new RemoveNthNodeFromEndOfList().new Solution();
    }
    public class ListNode {
        int val;
        ListNode next;
        ListNode() {}
        ListNode(int val) { this.val = val; }
        ListNode(int val, ListNode next) { this.val = val; this.next = next; }
  }
    //leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
       /* if (head == null || head.next==null) return null;
        int size = 1;
        ListNode temp = head.next;
        while (temp != null){
            size++;
            temp = temp.next;
        }

        if (size == n) return head.next;

        temp = head;
        while (temp!=null){
            size--;
            if (size == n){
                temp.next = temp.next.next;
                return head;
            }
            temp = temp.next;
        }
        return head;*/


        ListNode slow = head;
        ListNode fast = head;
        for (int i = 0; i < n; i++) {
            fast = fast.next;//保持节点的间距
        }
        if (fast == null) return head.next;//删除的节点是头结点
        while (fast.next != null){
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return head;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

}